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Let $p$ be an odd prime number and let $\alpha = [X] \in R=\mathbb F_p[X]/\langle X^4+1\rangle$, and $y = \alpha + \alpha^{-1}$

I've proven:

1) $\alpha$ is a primitive eight root of unity in $R$.

2) $y^2 = 2$ and $y^p = \alpha^p + \alpha^{-p}$.

$\alpha^{-p} = {(\alpha^{-1})}^{p}$ right ?

3) $y^p = y$ if $p \equiv 1,7 \pmod 8$ and $y^p = -y$ if $p\equiv 3,5 \pmod 8$.

However I need help proving that $\left(\frac2p\right) = 1$ if $p \equiv 1,7 \pmod 8$ and $\left(\frac2p\right) = -1$ if $p \equiv 3,5 \pmod 8$. This should be accomplished using what I've already proven according to the text.

Please dont forget comment on my "$\alpha^{-p} = {(\alpha^{-1})}^{p}$ right ?".

Thanks.

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  • $\begingroup$ I'd let $\pi(X)$ be a prime factor of $X^4+1$ and consider the ring $R=F_p[X]/\langle\pi(X)\rangle$, since this is a field, letting you do some simplifications. $\endgroup$ – Thomas Andrews Dec 5 '13 at 13:43
  • $\begingroup$ $R$ is indeed a field. Let $F$ be a field and $f \in F[X]$ an irreducible polynomial. Then $R=F[X]/\langle f \rangle$ is a field. $\endgroup$ – Shuzheng Dec 5 '13 at 13:43
  • $\begingroup$ No, $R$ is not necessarily a field. $X^4+1$ is not necessarily an irreducible polynomial. $\endgroup$ – Thomas Andrews Dec 5 '13 at 13:44
  • $\begingroup$ Ohh, guess you are right. But I really would like my answer to be based on 1,2,3. However since the problem assume $\alpha^{-1}$ exists doesn't that imply we are working in a field ? $\endgroup$ – Shuzheng Dec 5 '13 at 13:48
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Yes, right, $\alpha^{-p} = \left(\alpha^{-1}\right)^p$.

Regarding the Legendre symbol, note that

$$y^p = \left(y^2\right)^{(p-1)/2}\cdot y = 2^{(p-1)/2}\cdot y = \left(\frac{2}{p}\right)\cdot y.$$

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  • $\begingroup$ why $$2^{(p-1)/2}\cdot y = \left(\frac{2}{p}\right)\cdot y.$$ ? and could you tell me how we can assume $\alpha^{-1}$ exists, since we don't know if $R$ is a field. Thanks a lot btw. $\endgroup$ – Shuzheng Dec 5 '13 at 13:53
  • $\begingroup$ Is it because $$\left(\frac{2}{p}\right) \equiv 2^{(p-1)/2} (\mod p)$$ ? $\endgroup$ – Shuzheng Dec 5 '13 at 13:57
  • $\begingroup$ First $\alpha^{-1}$. The polynomial factored out is $X^4+1$, which means $\alpha^4 = -1$, and hence $\alpha^8 = 1$, so $\alpha^{-1} = \alpha^7$. Second, the Legendre symbol, yes, precisely, because $a^{(p-1)/2} \equiv \left(\frac{a}{p}\right) \pmod{p}$. $\endgroup$ – Daniel Fischer Dec 5 '13 at 13:59
  • $\begingroup$ I know $$\left(\frac{2}{p}\right) \equiv 2^{(p-1)/2}$$ but this just means that $\left(\frac{2}{p}\right) = 2^{(p-1)/2} + pk$. But then $$\left(\frac{2}{p}\right) \cdot y = (2^{(p-1)/2} + pk) \cdot y = 2^{(p-1)/2} \cdot y$$ why ? $\endgroup$ – Shuzheng Dec 5 '13 at 14:28
  • $\begingroup$ @user111854 We're working over $\mathbb{F}_p$, so all integers, like $2^{(p-1)/2}$ or $\left(\frac{2}{p}\right)$ are reduced modulo $p$/represent their residue class modulo $p$. In $\mathbb{F}_p$, we have the equality, $a^{(p-1)/2} = \left(\frac{a}{p}\right)$, not only the congruence. $\endgroup$ – Daniel Fischer Dec 5 '13 at 14:32

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